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a(n) = Sum_{k=0..n} Sum_{j=0..k} (binomial(n,k) * binomial(k,j))^n.
4

%I #12 Jul 17 2020 17:05:33

%S 1,3,15,381,67635,83118753,813824623689,58040410068847251,

%T 32150480245981639533315,154935057570894645075940703673,

%U 5474671509704049919709361235659936825,1600436120524545216094358662984789029130593831

%N a(n) = Sum_{k=0..n} Sum_{j=0..k} (binomial(n,k) * binomial(k,j))^n.

%H Seiichi Manyama, <a href="/A336270/b336270.txt">Table of n, a(n) for n = 0..47</a>

%F a(n) = (n!)^n * [x^n] (Sum_{k>=0} x^k / (k!)^n)^3.

%t Table[Sum[Sum[(Binomial[n, k] Binomial[k, j])^n, {j, 0, k}], {k, 0, n}], {n, 0, 11}]

%t Table[(n!)^n SeriesCoefficient[Sum[x^k/(k!)^n, {k, 0, n}]^3, {x, 0, n}], {n, 0, 11}]

%o (PARI) a(n) = sum(k=0, n, sum(j=0, k, (binomial(n,k) * binomial(k,j))^n)); \\ _Michel Marcus_, Jul 16 2020

%Y Cf. A000244, A002893, A141057, A167010, A172434, A180350.

%K nonn

%O 0,2

%A _Ilya Gutkovskiy_, Jul 15 2020